Answer:
8 N
Explanation:
Using the equation F=ma (F: force/ m: mass in kg/ a: acceleration),
F = (800/1000)(10)
F = 8 N
Answer:
v = 10 m/s
Explanation:
Let's assume the wheel does not slip as it accelerates.
Energy theory is more straightforward than kinematics in my opinion.
Work done on the wheel
W = Fd = 45(12) = 540 J
Some is converted to potential energy
PE = mgh = 4(9.8)12sin30 = 235.2 J
As there is no friction mentioned, the remainder is kinetic energy
KE = 540 - 235.2 = 304.8 J
KE = ½mv² + ½Iω²
ω = v/R
KE = ½mv² + ½I(v/R)² = ½(m + I/R²)v²
v = √(2KE / (m + I/R²))
v = √(2(304.8) / (4 + 0.5/0.5²)) = √101.6
v = 10.07968...
Answer:
Explanation:
Given that,
Frequency of radio signal is
f = 800kHz = 800,000 Hz.
Distance from transmitter
d = 8.5km = 8500m
Electric field amplitude
E = 0.9 V/m
The average energy density can be calculated using
U_E = ½•ϵo•E²
Where ϵo = 8.85 × 10^-12 F/m
Then,
U_E = ½ × 8.85 × 10^-12 × 0.9²
U_E = 3.58 × 10^-12 J/m²
The average electromagnetic energy density is 3.58 × 10^-12 J/m²
The question is incomplete. I can help you by adding the information missing. They want you to calculate a) the radius of the cyclotron orbit for an electron with speed 1.0 * 10^6 m/s^2 and b) the radius of a cyclotron orbit for a proton with speed 5.0 * 10^4 m/s.
The two tasks involve combining the equations of the magnectic force and the centripetal force in a circular motion.
When you do that, you will obtain an expression to find the radius of the circular motion, which is the radius of the cyclotron that impulses the particles.
a)
Magentic force, F = q*v*B
q is the charge of the electron = 1.6 * 10^ -19 C
v is the speed = 1.0 * 10 ^ 6 m/s
B is the magentic field = 5.0 * 10 ^-5 T
Centripetal force, F = m*Ac = m * v^2 / R
where,
Ac = centripetal acceleration
m = mass of the electron = 9.11 * 10 ^-31 kg
R = the radius of the orbit
Now equal the two forces: q*v*B = m * v^2 / R => R = m*v / (q*B)
=> R = (9.11 * 10^31 kg) (1.0*10^6m/s) / [ (1.6 * 10^-19C)* (5.0 * 10^-5T) ]
=> R = 0.114 m
b) The equations are the same, just now use the speed, charge and mass of the proton instead of those of the electron.
R = m*v / (qB) = (1.66*10^-27 kg)(5.0*10^4 m/s) / [(1.6*10^-19C)(5*10^-5T)]
=> R = 10.4 m
This study was aimed at testing the construct validity of the basketball basic motion skills test instrument (ITK GDBB). The research used descriptive method of 3 basketball experts in the city of Cimahi; 3 experts are the expert in basketball. The instrument used was the ITB GDBB developed by Silvy (2019) consisting of top passing, bottom passing, top service, bottom service, chest passing, bounding passing, overhead passing, and leading ball (dribbling). This instrument consists of 76 items that cover 4 domains in basketball, namely chest pass, overhead pass, bound pass, and dribbling. The validity method used the construct validity of different power types. For the reliability method, it used the Kuder Ricardson (KR) and Objectivity analysis. The results of the construct validity analysis of a total of 76 items show that the score is ranged from 0.67 to 1.00. The construct validity value of 71 items in the basketball game is in the high category (= 1.00), 5 items are in the sufficient category, the relativity score is ranged from 0.75 to 0.98, and the objectivity score is ranged from 0.89 to 0.95. The conclusion is that this test instrument can be used as a standardized basic motion skill test for standardized large ball games for validity in basic motion skills in basketball games for grade VII junior high school students.